Question: $\dfrac{ 9q + 9r }{ -6 } = \dfrac{ 3q - 7s }{ -5 }$ Solve for $q$.
Solution: Multiply both sides by the left denominator. $\dfrac{ 9q + 9r }{ -{6} } = \dfrac{ 3q - 7s }{ -5 }$ $-{6} \cdot \dfrac{ 9q + 9r }{ -{6} } = -{6} \cdot \dfrac{ 3q - 7s }{ -5 }$ $9q + 9r = -{6} \cdot \dfrac { 3q - 7s }{ -5 }$ Multiply both sides by the right denominator. $9q + 9r = -6 \cdot \dfrac{ 3q - 7s }{ -{5} }$ $-{5} \cdot \left( 9q + 9r \right) = -{5} \cdot -6 \cdot \dfrac{ 3q - 7s }{ -{5} }$ $-{5} \cdot \left( 9q + 9r \right) = -6 \cdot \left( 3q - 7s \right)$ Distribute both sides $-{5} \cdot \left( 9q + 9r \right) = -{6} \cdot \left( 3q - 7s \right)$ $-{45}q - {45}r = -{18}q + {42}s$ Combine $q$ terms on the left. $-{45q} - 45r = -{18q} + 42s$ $-{27q} - 45r = 42s$ Move the $r$ term to the right. $-27q - {45r} = 42s$ $-27q = 42s + {45r}$ Isolate $q$ by dividing both sides by its coefficient. $-{27}q = 42s + 45r$ $q = \dfrac{ 42s + 45r }{ -{27} }$ All of these terms are divisible by $3$ Divide by the common factor and swap signs so the denominator isn't negative. $q = \dfrac{ -{14}s - {15}r }{ {9} }$